# Proof

By hypothesis, $(X,\mathcal O)$ is a topological space and $B\subset X$ its subset.

### Ad $(1)$

• Let $x\in B \setminus \delta B$ be an element of the interior.
• Since $x$ is not a boundary point, there is a neighborhood $N_x$ that has no points in common with the complement $B^C$.
• Moreover, we can choose $N_x$ in a way that it has even no points in common with boundary $\delta B.$ Otherwise, if there existed an $y\in N_x \cap \delta B,$ then $N_x$ would be a neighborhood of $y$ and would have no points in common with the complement $B^C.$ But every neighborhood of a boundary point has points in common with the complement.
• Thus, $N_x$ has no boundary point in it.
• Altogether, we have shown that every neighborhood of $x$ is contained in the interior.
• It follows that the interior is open.

### Ad $(2)$

• Let $B^C$ be the complement of $B.$
• From the definition of the boundary it follows that $\delta B^C=\delta B$.
• From $(1)$ we can deduce $B^C\setminus \delta B^C)$ is open.
• Therefore $(B^C\setminus \delta B^C)^C$ is closed.
• But $(B^C\setminus \delta B^C)^C=B\cup \delta B,$ which is the closure.
• Therefore, the closure is closed.

### Ad $(3)$

• We have $\delta B=(B\cup \delta B)\setminus(B\setminus \delta B).$
• It follows that $X\setminus \delta B=(X\setminus(B\cup\delta B))\cup(B\setminus \delta B),$ which is, according to $$(1)$$ and $$(2)$$, a union of two open sets.
• By the third axiom of topology, $X\setminus \delta B=(\delta B)^C$ is open.
• Therefore, the complement $((\delta B)^C)^C)=\delta B$ is closed, which is the boundary.

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### References

#### Bibliography

1. Forster Otto: "Analysis 2, Differentialrechnung im $$\mathbb R^n$$, Gewöhnliche Differentialgleichungen", Vieweg Studium, 1984